In this example, the objective is to be able to withdraw $30,000 per year from savings. It assumes 7.5% interest (it's obviously very old!). The "end balance" is, of course,0 by definition.
Suppose you had X dollars in the bank, drawing 7.5% simple interest.
After 1 year, before you withdraw anything, you would have the original X plus interest, (0.075)(30000) or a total of (1+ 0.075)X= 1.075X. Now, you withdraw 30000. You have 1.075X- 30000 left.
For the second year, repeat that with initial amount 1.075X-30000. Before withdrawing anything, but including interest, you would have
1.075(1.075X- 30000)= (1.075)2X- (1.075)(30000). After withdrawing your 30000 you will have (1.075)2X- (1.075)(30000)- 30000 left.
For the third year, you are starting with that amount so, before withdrawal you would have 1.075((1.075)2X- 1.075(30000)- 30000)= 1.0753X- 1.0752300000- 1.075(30000). Now withdraw 30000 from that: 1.0753X- 1.0752300000- (1.075)30000- 30000= 1.0753X- 30000(1+ 1.075+ 1.0752.
Do you see the pattern? After n years you will have 1.075nX- (30000)(1+ 1.075+ ...+ 1.075n-1. In particular, after 5 years you would have 1.0755X- (30000)(1+ 1.075+ 1.0752+ 1.0753+ 1.0754. Since in this example you are apparently only expecting to die 5 years after retirement, after 5 years, the "final balance" is to be 0. It's not that hard to calculate that 1+ 1.075+ 1.0752+ 1.0753+ 1.0754= 5.808 approximately. Solve the equation 1.0755X= 30000(5.808). I get slightly less than 130480. Try deducting the 30000 before adding the interest. That would be the same as replacing X by X-30000.